What is the solution for y^{''}+2y^'+2y=0,y^0=0,y^'0=1 ?

The solution for y^{''}+2y^'+2y=0,y^0=0,y^'0=1 is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))

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y^{''}+2y^'+2y=0,y^0=0,y^'0=1

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y^{0} (0) = 0, y^{^{'}} 0 = 1

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t))

4 \frac{d ^{2} y}{d x ^{2}} + 32 \frac{d y}{d x} + 64 y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 8, y^{^{'}} (0) = 9

y^{^{''}} + 6 y = 0, y (0) = 0, y (1) = 0

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = a, y^{^{'}} (0) = - 40

77 y^{^{''''}} + 303 y^{^{''}} + 100 y = 0

What is the solution for y^{''}+2y^'+2y=0,y^0=0,y^'0=1 ?
The solution for y^{''}+2y^'+2y=0,y^0=0,y^'0=1 is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))